Solve 9x + 5 = 41
The solution of 9x + 5 = 41 is x = 4.
How to work it out
- 1
The goal is to get x alone. Undo the operations in the opposite order: first the addition, then the multiplication.
- 2
Subtract 5 from both sides:
- 3
Divide both sides by 9:
- 4
Check: put x = 4 back into the equation.
LaTeX source of the formulas on this page
\[ 9x + 5 = 41\ \Rightarrow\ x = 4 \]
\[ 9x + 5 = 41 \]
\[ 9x + 5 - 5 = 41 - 5 \ \Rightarrow\ 9x = 36 \]
\[ \frac{9x}{9} = \frac{36}{9} \ \Rightarrow\ x = 4 \]
\[ 9\cdot4 + 5 = 36 + 5 = 41 \checkmark \]Trying values of x in 9x + 5 = 41
| x | 9x + 5 | Equals 41? |
|---|---|---|
| 2 | 23 | no |
| 3 | 32 | no |
| 4 | 41 | yes |
| 5 | 50 | no |
| 6 | 59 | no |
Questions
What is x in 9x + 5 = 41?
The solution of 9x + 5 = 41 is x = 4.
How do you solve 9x + 5 = 41?
Subtract 5 from both sides to get 9x = 36, then divide both sides by 9 to get x = 4.
How do I check the answer?
Replace x by 4: 9(4) + 5 = 41, which is 41.